Question

Differentiate \[\sin \left( 2 \sin^{- 1} x \right)\] ?

Answer 1

\[\text{ Let } y = \sin\left( 2 \sin^{- 1} x \right)\]

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left[ \sin\left( 2 \sin^{- 1} x \right) \right]\]

\[ = \cos\left( 2 \sin^{- 1} x \right)\frac{d}{dx}\left( 2 \sin^{- 1} x \right) \left[ \text{Using chain rule} \right]\]

\[ = \cos\left( 2 \sin^{- 1} x \right) \times 2\frac{1}{\sqrt{1 - x^2}}\]

\[ = \frac{2\cos\left( 2 \sin^{- 1} x \right)}{\sqrt{1 - x^2}}\]

\[So, \frac{d}{dx}\left[ \sin\left( 2 \sin^{- 1} x \right) \right] = \frac{2\cos\left( 2 \sin^{- 1} x \right)}{\sqrt{1 - x^2}}\]